
Functions
Hello :)
The problem states:
IF $\displaystyle f(\frac{1}{x})=x^2+\frac{1}{x^3}$ AND $\displaystyle g(x)= \frac{1}{2}(f(x)+f(x))$ THEN $\displaystyle g(x)=?$
So, I got that $\displaystyle f(x)=\frac{1}{x^2}+\frac{x^3}{1}$ and that $\displaystyle f(x)=\frac{1}{x^2}\frac{x^3}{1}$
That gives me $\displaystyle g(x)=\frac{1}{2}*\frac{2}{x^2}=\frac{1}{x^2}$
Is this correct? I had huge problems with this as I was always getting $\displaystyle g(x)=\frac{1}{2}*0$... then I figured my mistake: $\displaystyle f(x)$ is not equal to $\displaystyle (f(x))$

"Is this correct?" Yes, it is.

If f is any function, we can define the "even part" of f(x) to be $\displaystyle f_e(x)= \frac{f(x)+ f(x)}{2}$ and the "odd part" of f(x) to be [tex]f_o(x)= \frac{f(x) f(x)}{2}[/itex]. $\displaystyle f_e$ is an "even function", f(x)= f(x), $\displaystyle f_o(x)$ is an "odd functin", f(x)= f(x), and their sum is $\displaystyle f_e(x)+ f_o(x)= f(x)$.
If f is an even function to start with, its "odd part" will be 0 and if f is an odd function, its "even part" will be 0. Rational functions that involve only even powers of x are "even functions" and rational functions that involve only odd powers of x are "odd functions.
Here, you were asked to find the even part of the function $\displaystyle f(x)= \frac{1}{x^2}+ x^3$ which is, as you finally found, $\displaystyle \frac{1}{x^2}$. Its odd part is, of course, $\displaystyle x^3$.
Here, you are basically asked